All intermediate sub extensions of $\mathbb{Q} \subseteq K \subseteq \mathbb{Q}(\zeta_8)$.

Question

I know there is a similar question posted on Stack Exchange, however it deals with periods, and I do not understand the solutions provided.

I know that the Galois Group of the field extension $\mathbb{Q}(\zeta_8)/\mathbb{Q}$ is $\mathbb{Z}_2 \times \mathbb{Z}_2$. To find the intermediate field extensions, we look at the subgroups of $\mathbb{Z}_2 \times \mathbb{Z}_2$. These are: $\{ (0,0) \}, \{(0,0), (0,1) \}, \{(0,0), (1,0) \}, \{(0,0), (1,1) \}$ and $\{(0,0), (0,1), (1,0), (1,1) \}$. Furthermore, I'm aware that $\zeta_8$ has minimal polynomial $x^4+1$, which has roots $\pm \zeta_4$ and $\pm \zeta_8$. I need help determining the explicit correspondence between the subgroups and sub extensions.

Answer 1

You can do this ad hoc. Note we know all the automorphisms already

$$\begin{cases} \zeta_8\mapsto\zeta_8 \\ \zeta_8\mapsto \zeta_8^{-1} \\ \zeta_8\mapsto \zeta_8^{3} \\ \zeta_8\mapsto\zeta_8^5\end{cases}.$$

The first one is the identity which fixes the whole field, so that's not our issue and the whole group clearly fixes the base field so that's not an interesting case either. So let us consider the three quadratic subfields.

We see $\zeta_8+\zeta_8^{-1}$ is fixed by complex conjugation, $\zeta_8\mapsto\zeta_8^{-1}$, but not $\zeta_8\mapsto \zeta_8^3$ or $\zeta_8\mapsto\zeta_8^5$ so it is the fixed field of that automorphism. Similarly $\zeta_8+\zeta_8^3$ is fixed not by complex conjugation, but is fixed by $\zeta_8+\zeta_8^3$. Finally $\zeta_8^2=\zeta_4$ is fixed by $\zeta_8\mapsto\zeta_8^5$ but conjugation makes this $\zeta_8^{6}$ and the second automorphism makes it $\zeta_8^{6}$ as well, neither of which fix it, so this generates the third fixed field. But we can do better.

• For the first one $(\zeta_8+\zeta_8^{-1})^2=\zeta_4+\zeta_4^{-1}+2=2$ so the fixed field of conjugation is just $\Bbb Q(\sqrt 2)$, which is what you expect.
• The second one is $\zeta_8(1+i)=i\sqrt 2$ by factoring out a $\zeta_8$ because squaring this gives $i\cdot 2i=-2$ and so this fixed field is $\Bbb Q(i\sqrt 2).$
• Similarly $\zeta_8^2=\zeta_4$ so the fixed field of the third automorphism is just $\Bbb Q(i)$.